Let $a(x)=-x^4$, and $b(x)=x^2+1$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Answer: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{-x^4}{x^2+1}$ : First, we divide ${x^2}$ into ${-x^4}$ and get ${-x^2}$ : $ \hphantom{1567|1} {-x^2}\\ {{{x^2}+1}}|\overline{{-x^4}+0x^2}\\ \hphantom{37.....|}\llap{-}\underline{(-x^4-x^2)}\\ \hphantom{37|3............}+x^2\\ $ [What did we do here?] Next, we divide ${x^2}$ into ${x^2}$ to get ${+1}$ : $ \hphantom{1567|} {-x^2 {+\ 1}}\\ {{{x^2}+1}}|\overline{x^4+0x^2+0}\\ \hphantom{37...|}\llap{-}\underline{(-x^4-x^2)}\\ \hphantom{37|3..........}+x^2+0\\ \hphantom{37...............|}\llap{-}\underline{(x^2+1)}\\ \hphantom{37|3...................}{-1}\\ $ [What did we do here?] The process stops here because $x^2+1$ is a polynomial of the second degree and $-1$ is a polynomial of the zeroth degree. So it follows that ${r(x)}={-1}$, ${q(x)}={-x^2+1}$, and $ \dfrac{-x^4}{x^2+1}={-x^2+1}+\dfrac{{-1}}{x^2+1}$ To conclude, $q(x)=-x^2+1$ $r(x)=-1$